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Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}\ln x,\,x > 0} \\
{0,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0}
\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $
$-2$
$-1$
$0$
$0.5$
Solution
(d) For Rolle’s theorem to be applicable to $f,$ for $x \in [0,\,1]$, we should have
$(i)$ $f(1) = f(0)$,
$(ii)$ $f$ is continuous for $x \in [0,\,1]$ and $f$ is differentiable for $x \in (0,\,1)$
From $(i),$ $f(1) = 0$, which is true.
From $(ii),$ $0 = f(0) = f({0_ + }) = \mathop {\lim }\limits_{x \to {0_ + }} {x^\alpha }\ln x$
Which is true only for positive values of $\alpha $, thus $(d)$ is correct.
Similar Questions
Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:
$x=-1$ | $x=0$ | $x=2$ | |
$f(x)$ | $3$ | $6$ | $0$ |
$g(x)$ | $0$ | $1$ | $-1$ |
In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)
$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$
$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$